Banging my head against monoids

Another problem from my category theory book, section 4.6 exercise 3: Show that M is a group if and only if ∅ and M are the only left ideals.

No arrow drawing this time! This is pretty much straight algebra. The first half I found easy.

Unpacking the definitions: a monoid has an identity element (1) and an associative operator (*). A group is a monoid where every element 'a' has an inverse 'a-1' such that a * a-1 = 1. A left ideal is a subset of the monoid that is closed under "left-multiplication", i.e., if B is a left-ideal, then m*b is in B for any b in the ideal and any m in the monoid.

Ready? Go!

(=>) Suppose M is a group, and suppose B is a nonempty left-ideal (there must be at least one, M itself.) Pick some element b from B. Now, b-1 is in M, so b-1 * b = 1 must be in B. But once the identity is in B, so must everything else in M, so B=M. Thus M has only itself and the empty set as ideals.

(<=) Suppose M has just two left ideals.

Now I hit a roadblock. How the heck could we get from only two left ideals, back to the existence of inverses? I mean, monoids include all sorts of crazy-ass stuff; who was I to judge whether they were allowed more than two ideals? So, time to look at some examples:

Say M=(N,+,0), the natural numbers under addition. Their left ideals are N itself, but also N-{0}, N-{0,1}, N-{0,1,2} and so on. If we take B={2,3,4,5,6...} then adding anything in N only gives us a number 2 or higher.

Or let's say M is a set of clockwise rotations. Usually rotations have inverses--- two 180 degree rotations cancel out. But say we can rotate by 30, 45, and sqrt(2) degrees. Then the monoid that gets generated has inverses for all of the rational-degree rotations, but none of the irrational-degree rotations. The irrational-degree rotations form a left ideal.

So from these examples it appears that the non-inverse elements generate left ideals. The element 2 gives us the left-ideal 2+N. The element sqrt(2) gives us the left-ideal of rotations by sqrt(2)*15*k degrees, k>=1. This might let us demonstrate a specific ideal when M is *not* a group, so I tried the logically equivalent:

(<=) Suppose M is not a group. (want to show: M has more than two left ideals.)

If M is not a group, at least one element a does not have an inverse. Consider B=M*a (m*a for all m in M.)

Fix some arbitrary element b from B, b=m*a for some m in M, by our definition. Then for any n in M, n*b = n*(m*a). But n*m is also in M, so (n*m)*a is in B too. (Yay for associativity!) Thus B is a left ideal.

But, B can't contain the identity element, because a-1 doesn't exist in M. So B is not equal to M, and B's nonempty too. Thus we have at least one additional left-ideal.

(Relevance to Goldblatt: for a given monoid M, the M-sets and their equivariant functions constitute a category which is a topos. But the 'truth values' of this topos are all left ideals of M.)